Why can we substitute $ V_{\mu \nu} $ to $ V_{\mu ; \nu} $ while inducing contracted Bianchi identity?

Question

After $ ( A_\mu B_\nu )_{; \sigma ; \rho} - ( A_\mu B_\nu )_{; \rho ; \sigma} = A_\alpha B_\nu R^\alpha_{\mu \rho \sigma} + A_\mu B_\alpha R^\alpha_{\nu \rho \sigma} $ where $ A_\mu B_\nu $ is outer product of two vector $ A_\mu $ and $ B_\nu $, I understand until substituting $ A_\mu B_\nu $ to $ V_{\mu \nu} $ so it becomes $ V_{\mu \nu ; \sigma ; \rho} - V_{\mu \nu ; \rho ; \sigma} = V_{\alpha \nu} R^\alpha_{\mu \rho \sigma} + V_{\mu \alpha} R^\alpha_{\nu \rho \sigma} $. But I don't get the reason why we can substitute $ V_{\mu \nu} $ to $ V_{\mu ; \nu} $ so we get $ V_{\mu ; \sigma ; \rho ; \nu} - V_{\mu ; \rho ; \sigma ; \nu} = V_{\alpha ; \nu} R^\alpha_{\mu \rho \sigma} + V_{\mu ; \alpha} R^\alpha_{\nu \rho \sigma} $. Isn't $ V_{\mu \nu} $ a rank-2 tensor and $ V_{\mu ; \nu} $ a vector?

Answer 1

The formula $$ T_{\mu \nu ; \sigma ; \rho} - T_{\mu \nu ; \rho ; \sigma} = T_{\alpha \nu} R^\alpha_{\mu \rho \sigma} + T_{\mu \alpha} R^\alpha_{\nu \rho \sigma} $$ holds for all rank-2 tensors $T_{\mu \nu}$. (And $T_{\mu \nu}$ does not have to be of the form $A_\mu B_\nu$.)

In particular, $V_{\mu ; \nu}$ is a rank-2 tensor, so the formula applies to it.